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We have a call option on 3 dices with strike 3. What's the fair value of the call when it pays the min value of the 3 dices?

E.g if we throw and have 426, the min is 2 here and so call is OTM (S < K)

216 permutations (6^3)

min(6) = 1/216 x (6-3) = 0.0139 EV

(666)

min(5) = 8/215 x (5-3) = 0.0744 EV

(555 556 565 566 655 656 665 666)

min(4) = ?

I'm getting lost in the permutation formula to calculate the number of outcomes where the min. comes 4. Can someone help me figure out how to find out the correct # of outcomes with value 4?

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  • $\begingroup$ 666 does not yield a minimum of 5, fyi. $\endgroup$ – Attack68♦ Feb 4 '20 at 17:20
  • $\begingroup$ @Attack68 No but an option with a strike would be ITM if three sixes are rolled. $\endgroup$ – D Stanley Feb 4 '20 at 18:49
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There is a simple way to find the number of rolls with the minimum of 4:
the number of rolls with the minimum of 6 is 1.
The number of rolls with the minimum of 5 is the number of rolls for which all outcomes are 5 or 6 minus the number of rolls with the minimum of 6:
2*2*2-1=7.
The number of rolls with the minimum of 4 is the number of rolls for which all outcomes are 4,5,6 minus the number of rolls with the minimum of 5 or 6:
3*3*3-7-1=19 and so on.

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  • $\begingroup$ Thank you, That's perfect! $\endgroup$ – aallove Feb 4 '20 at 20:57
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This isn't really an option as it is an exercise in probability.

How many rolls have a minimum of 6?

1 = 3C3 (6s)

How many rolls have a minimum of 5?

7 = 3C3 (5s) + 3C2 (5s6s) + 3C1 (5s6s)

How many rolls have a minimum of 4?

19 = 3C3 (4s) + 3C2 (4s5s) + 3C1 (4s5s) + 3C2 (4s6s) + 3C1 (4s6s) + 3P3 (4s5s6s)

How many rolls have a minimum of 3?

  3C3 (3s)
  3C2 (3s4s) 3C2 (3s5s) 3C2 (3s6s)
  3C1 (3s4s) 3C1 (3s5s) 3C1 (3s6s)
  3P3 (3s4s5s) 3P3 (3s5s6s) 3P3 (3s4s6s) = 3P3 * 3C2

= 1 + 9 + 9 + 18 = 37

How many rolls have minimum of 2?

 3C3 (2s)   = 1
 3C2 * 4C1  = 12
 3C1 * 4C1  = 12
 3P3 * 4C2  = 36

= 61

How many rolls have minimum of 1?

3C3    = 1
3C2 * 5C1 = 15
3C1 * 5C1 = 15
3P3 * 5C2 = 60

= 91

Total is 216

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  • $\begingroup$ Thanks for the breakdown! $\endgroup$ – aallove Feb 4 '20 at 20:58

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